# Irreducible polynomial in non-UFD

I'm asked to proved that $x^n-11$ is irreducible in $\mathbb{Q}(\sqrt{-5})[x]$. I have deduced to a point where I have to prove that if $x^n-11$ is reducible in $\mathbb{Q}(\sqrt{-5})[x]$ then it is also reducible in $\mathbb{Z}[\sqrt{-5}][x]$.

Is this fact true? If it is, can someone tell me how to prove that. If not, how else should I approach the original question.

Note that $\mathbb{Z}[\sqrt{-5}][x]$ is not a Unique Factorization Domain! Thanks!

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11 is prime in $\mathbb{Z}(\sqrt{-5})$ – user44322 Dec 3 '12 at 8:16
To invoke Eisenstein, you must first use Gauss's Lemma to work in $\mathbb{Z}[\sqrt{-5}][x]$ but you cant use Gauss's Lemma in non-UFD – user44322 Dec 3 '12 at 8:22
I am completely wrong. Sorry. Since $\mathbb{Z}[\sqrt{-5}]$ is not a UFD, you cannot use irreducibility in $\mathbb{Z}[\sqrt{-5}]$ to show irreducibility in $\mathbb{Q}(\sqrt{-5})$. I deleted my earlier comments. Please see Zach's answer below. It is correct. – Rankeya Dec 3 '12 at 8:22

(My algebra is not great, so please check this carefully.) We have $[\mathbb{Q}(\sqrt{-5}):\mathbb{Q}] = 2$. Now, $x^n - 11$ is irreducible over $\mathbb{Q}$ by Eisenstein, and so $[\mathbb{Q}(\sqrt[n]{11}):\mathbb{Q}] = n.$ Since $\mathbb{Q}(\sqrt[n]{11})$ is a subfield of $\mathbb{R}$, it follows that $\sqrt{-5}$ is not contained in it. Hence $[\mathbb{Q}(\sqrt{-5},\sqrt[n]{11}):\mathbb{Q}(\sqrt[n]{11})] = 2.$ But then $[\mathbb{Q}(\sqrt{-5},\sqrt[n]{11}):\mathbb{Q}(\sqrt{-5})] = n.$ Hence the irreducible polynomial of $\sqrt[n]{11}$ over $\mathbb{Q}(\sqrt{-5})$ must have degree $n$, and so is $x^n-11$. In particular, that polynomial is irreducible.
Dear @ZachL.: $[\mathbb{Q}(11^{1/n}):\mathbb{Q}] = n$. – Rankeya Dec 3 '12 at 8:20
$[\mathbb{Q}(\sqrt{-5},\sqrt[n]{11}):\mathbb{Q}(\sqrt{-5})] = n$ Why is this true? – user44322 Dec 3 '12 at 8:32